Recursive sequence $a_{n+1} = \sqrt{a_1 + … + a_n}$ [duplicate]

I need a hint to solve this:

Let $a_1 = 1$ and define a sequence recursively by $$a_{n+1} = \sqrt{a_1 + a_2 + ... a_n}$$

Show that $$\lim_{n \to \infty} \dfrac{a_n}{n} = \dfrac{1}{2}$$

Any help? Thank you

marked as duplicate by Sil, Xander Henderson, max_zorn, Namaste, Adrian KeisterAug 12 '18 at 1:06

• Another way of writing the recursion is as follows:$$a_{n+1}=\sqrt{a_n^2+a_n}$$ – Akiva Weinberger Oct 2 '14 at 3:09
• Actually that was my 1st try, but I couldn't finish it. Then I used that to get $a_{n+1} \le \sqrt{2} a_n$ but nothing. – FormerMath Oct 2 '14 at 3:12
• Use columbus8myhw's hint to prove that $a_n$ increases without bound. Then use the same relation to prove that $a_{n+1} = a_n + 1/2 + O(1/a_n)$. – user180040 Oct 2 '14 at 3:17
• Idea: If $a_n\le\frac n2$, then $a_{n+1}=\sqrt{a_n^2+a_n}\le\sqrt{\frac{n^2}4+\frac n2}=\frac{\sqrt{n^2+2n}}2<\frac{\sqrt{n^2+2n+1}}2=\frac{n+1}2$. Since $a_2=1<\frac22$, we know that (for $n\ge2$) $a_n\le\frac n2$. – Akiva Weinberger Oct 2 '14 at 3:24
First, you need to prove that $a_n\to\infty$ as $n\to\infty$. Then, by applying Stolz–Cesàro theorem we have $$\lim_{n\to\infty}\frac{a_n}{n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)-n}=\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}(\sqrt{a_{n}^2+a_n}-a_n)=\lim_{n\to\infty}\frac{a_n}{\sqrt{a_{n}^2+a_n}+a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{a_n}}+1}=\frac12.$$